Unit 6 · Topic 6.8 · about 30 minutes · 7 parts · 13 scored checks, 3 of them multiple choice

AP Calculus AB 6.8: Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation

Q: Why does every indefinite integral need + C?

Because a function has infinitely many antiderivatives, one for each constant. x2, x2 + 1 and x2 - 9 all have derivative 2x, and 2x\,dx = x2 + C names all of them.

Q: What is the difference between f(x)\,dx and ab f(x)\,dx?

The first, with no limits, is a family of functions, F(x) + C. The second, with limits, is a single number. Antiderivatives connect them: when f is continuous on [a, b], the Fundamental Theorem of Calculus gives ab f(x)\,dx = F(b) - F(a).

Q: Why is it |x| and not x?

x is only defined for x > 0, but 1x is also defined for x < 0. For x < 0 the chain rule gives ddx(-x) = 1x, so |x| covers both sides.

Q: Is there a product rule or a quotient rule for antiderivatives?

Not one you can apply term by term. Rewrite first when you can, by expanding a product or splitting a fraction into powers, and check the result by differentiating. Topic 6.9 adds a method that reverses the chain rule.

Q: If I can't find an antiderivative, does that mean there isn't one?

No. Some integrands, such as 2xex2, look hard and have a closed form. Others, such as ex2, have antiderivatives with no closed form. Either way, a continuous function on an interval always has an antiderivative.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 6: 15 to 20% of the multiple-choice section

What you'll be able to do

Find antiderivatives and indefinite integrals by running the derivative rules backward, and write them with integral notation and $+ C$ .

Key vocabulary

Term	What it means	The exam trap
Antiderivative	A function $F$ is an antiderivative of $f$ when $F^{'} (x) = f (x)$ .	Check a proposed antiderivative by differentiating it and comparing with the original function, not by how it looks.
Indefinite integral, $\int f (x) d x$	The family of all antiderivatives of $f$ , written $\int f (x) d x = F (x) + C$ , where $F^{'} (x) = f (x)$ and $C$ is any constant.	An indefinite integral is a family of functions, not a number; read whether an integral sign has limits before you answer.
Constant of integration, $C$	The arbitrary constant in $F (x) + C$ ; each value of $C$ gives a different antiderivative of the same function.	Write $+ C$ on every indefinite integral, and keep it to the last line of the work.
Closed-form antiderivative	An antiderivative written with finitely many familiar functions, such as powers, roots, exponentials, logarithms and trigonometric functions.	A function can have an antiderivative that has no closed form; failing to find a formula does not mean it has no antiderivative.

Graphs of y = x cubed over 3 minus x, plus 2, plus 0 and minus 2: three antiderivatives of x squared minus 1, each the same curve shifted up or down. At x = 2 a short black segment on each curve has slope 3.

Part 1 Think first How far after five seconds?

Commit to a guess first. Even a wrong guess makes the lesson stick better.

Predict first

A cyclist starts from rest and speeds up steadily, so that $t$ seconds later her speed is $2 t$ meters per second. How far has she ridden after 5 seconds?

In short. An antiderivative of $f$ is a function $F$ with $F^{'} (x) = f (x)$ , and the indefinite integral $\int f (x) d x = F (x) + C$ names every one of them at once.

Two antiderivatives of the same function on an interval differ by a constant, which is why an indefinite integral ends in $+ C$ .
Each derivative rule read backward is an antiderivative rule, so you check an antiderivative by differentiating it.
Some functions have antiderivatives that cannot be written in closed form, and those are handled numerically or with an accumulation function.

Part 2 Learn it Antiderivatives, one idea at a time

2a. An antiderivative, and the + C

Up to now you have gone from a function to its derivative. Here you go the other way. A function $F$ is an antiderivative of $f$ when $F^{'} (x) = f (x)$ .

Take $f (x) = 3 x^{2}$ . Because $\frac{d}{d x} (x^{3}) = 3 x^{2}$ , the function $x^{3}$ is an antiderivative of $f$ . So are $x^{3} + 4$ and $x^{3} - 7$ : a constant has derivative 0, so adding one does not change any slope. Their graphs are one curve shifted up or down, like the three curves in the picture at the top.

On an interval those shifts are the only antiderivatives there are, because two functions with the same derivative differ by a constant. The indefinite integral names the whole family at once:

\int 3 x^{2} d x = x^{3} + C,

where $C$ is any constant. In general $\int f (x) d x = F (x) + C$ , where $F^{'} (x) = f (x)$ . The function inside, $f (x)$ , is the integrand, and $d x$ says that $x$ is the variable.

An indefinite integral is a family of functions, not a number. The definite integral $\int_{a}^{b} f (x) d x$ from earlier in this unit is the number. To check an antiderivative, differentiate it and compare with $f$ .

Quick check · Decide and justify

No calculator. Let $f (x) = 2 x + 5$ . Is $G (x) = x^{2} + 5 x - 12$ an antiderivative of $f$ ?

Which reason supports your answer?

2b. Run the derivative rules backward

Every derivative rule you know, read from right to left, is an antiderivative rule. Each row below starts from a derivative you already have.

Derivative rules read backward
Because	the indefinite integral is
$\frac{d}{d x} (\frac{x^{n + 1}}{n + 1}) = x^{n}$ , for $n \neq - 1$	$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$
$\frac{d}{d x} \ln \| x \| = \frac{1}{x}$	$\int \frac{1}{x} d x = \ln \| x \| + C$
$\frac{d}{d x} e^{x} = e^{x}$	$\int e^{x} d x = e^{x} + C$
$\frac{d}{d x} \sin x = \cos x$	$\int \cos x d x = \sin x + C$
$\frac{d}{d x} (- \cos x) = \sin x$	$\int \sin x d x = - \cos x + C$
$\frac{d}{d x} \tan x = \sec^{2} x$	$\int \sec^{2} x d x = \tan x + C$

The power rule has a gap at $n = - 1$ , where it would divide by 0. That gap is filled by the logarithm row: $\frac{1}{x}$ gets $\ln | x |$ .

The derivative rules for sums and constant multiples run backward too: $\int k f (x) d x = k \int f (x) d x$ , and the integral of a sum is the sum of the integrals. So

\int (8 x^{3} + 2 \cos x) d x = 8 \cdot \frac{x^{4}}{4} + 2 \sin x + C = 2 x^{4} + 2 \sin x + C .

Rewrite roots and fractions as powers before you start: $\sqrt{x} = x^{1 / 2}$ and $\frac{1}{x^{3}} = x^{- 3}$ .

Quick check · Match

No calculator. Tap an integral, then the indefinite integral it equals.

2c. Some antiderivatives have no formula

A function that is continuous on an interval always has an antiderivative there. Whether you can write one down is a different question.

An antiderivative is in closed form when it is built from finitely many familiar functions (powers and roots, exponentials, logarithms, trigonometric and inverse trigonometric functions) by adding, subtracting, multiplying, dividing and composing them. Many functions have no closed-form antiderivative. Three you will meet are

e^{x^{2}}, \sin (x^{2}), \sqrt{1 + x^{3}} .

This is not a matter of nobody being clever enough: mathematicians have proved that no such formula exists.

Looks can mislead. $e^{x^{2}}$ has no closed-form antiderivative, but $2 x e^{x^{2}}$ has one, $e^{x^{2}}$ itself, because the chain rule gives $\frac{d}{d x} e^{x^{2}} = 2 x e^{x^{2}}$ . Before you decide that a function has none, ask whether it is the derivative of something you know.

A function with no closed-form antiderivative can still be worked with. A graphing calculator evaluates a definite integral of it numerically, and an accumulation function such as $\int_{0}^{x} e^{t^{2}} d t$ is an antiderivative of $e^{x^{2}}$ that needs no formula.

Quick check · Multiple choice

No calculator. Which of these functions has no closed-form antiderivative?

Common mistakes

The mistake	Why it is wrong	Do this instead
Leaving $+ C$ off an indefinite integral.	$\int f (x) d x$ stands for every antiderivative of $f$ . Without $+ C$ you have named only one of them.	End every indefinite integral with $+ C$ , and carry it to the last line of your work.
Writing $\int \sin x d x = \cos x + C$ .	The derivative of $\cos x$ is $- \sin x$ , so $\cos x$ is an antiderivative of $- \sin x$ , not of $\sin x$ .	Differentiate your answer before you move on. $\frac{d}{d x} (- \cos x) = \sin x$ , so $\int \sin x d x = - \cos x + C$ .
Using the power rule on $\frac{1}{x}$ , or on a power of something other than $x$ , such as $(1 + x^{3})^{1 / 2}$ .	The power rule undoes $\frac{d}{d x} x^{n + 1}$ for $n \neq - 1$ only. At $n = - 1$ it divides by 0, and on $(1 + x^{3})^{1 / 2}$ it ignores the chain rule factor.	Use $\ln \| x \|$ for $\frac{1}{x}$ . For a power of an inside function, check by differentiating; often there is no closed form at all.

Part 3 Watch it done An indefinite integral, term by term

Open each step in turn, then finish the last one yourself.

Find $\int (6 x^{2} - \frac{8}{x^{3}} + 4 \sin x) d x$ .

Step 1 · Rewrite as powers
$\frac{8}{x^{3}} = 8 x^{- 3}$ , so the integrand is $6 x^{2} - 8 x^{- 3} + 4 \sin x$ .
Step 2 · Split the integral
By the sum and constant multiple rules, $\int (6 x^{2} - 8 x^{- 3} + 4 \sin x) d x = 6 \int x^{2} d x - 8 \int x^{- 3} d x + 4 \int \sin x d x$ .
Step 3 · Reverse the power rule
$6 \int x^{2} d x = 6 \cdot \frac{x^{3}}{3} = 2 x^{3}$ and $- 8 \int x^{- 3} d x = - 8 \cdot \frac{x^{- 2}}{- 2} = 4 x^{- 2}$ , each up to a constant.
Why this step?
Why is the second term $+ 4 x^{- 2}$ when the integrand had $- 8 x^{- 3}$ ?
Raising the exponent from $- 3$ gives $- 2$ , and the rule divides by that new exponent. Dividing $- 8$ by $- 2$ gives $+ 4$ . A quick check: $\frac{d}{d x} (4 x^{- 2}) = - 8 x^{- 3}$ .
Your step · Fill the statement
No calculator. Finish it. Tap a blank, then tap an answer.

The derivative of is $- \sin x$ , so an antiderivative of $4 \sin x$ is .

Check the whole answer by differentiating: $\frac{d}{d x} (2 x^{3} + 4 x^{- 2} - 4 \cos x + C) = 6 x^{2} - 8 x^{- 3} + 4 \sin x$ . That is the integrand, so the antiderivative is right. One $+ C$ at the end covers the constants from all three terms.

Part 4 Try it New integrands

Try it · Sort it

No calculator. How would you find an antiderivative of each function, if you can write one at all? Tap a card, then a bin.

Reverse the power rule

Reverse a trigonometric derivative

No closed-form antiderivative

Try it · Calculate

Graphing calculator required. The function $\sqrt{1 + x^{3}}$ has no closed-form antiderivative, so $\int_{0}^{2} \sqrt{1 + x^{3}} d x$ cannot be found from an antiderivative by hand. Use your calculator to evaluate it. Give your answer to three decimal places.

Your answer

Part 5 Explain it Write it out, then score it

Free response on the exam is written by hand, so answer on paper first. Nothing you type on this page leaves it. Then open the model answer and give yourself each point you earned.

Free response is 50% of the exam score. This one is shaped like Question 4, from free-response Part B (calculator not permitted).

Free-response practice · Question 4 style

No calculator. Water is pumped into a storage tank at a rate modeled by $R (t) = 12 \sqrt{t} - 3 t$ gallons per hour, for $0 \leq t \leq 16$ , where $t$ is measured in hours.

(a) Find $\int R (t) d t$ . Show the antiderivative of each term.

(b) Verify your answer to part (a) by differentiating it.

(c) Let $W (t)$ be the number of gallons of water in the tank at time $t$ , so $W^{'} (t) = R (t)$ . Ana and Ben each model $W (t)$ with your answer to part (a); Ana takes $C = 200$ and Ben takes $C = 350$ . Explain why both models agree with the rate $R (t)$ . Using correct units, interpret the difference between the two models in the context of the problem.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) Write $12 \sqrt{t}$ as $12 t^{1 / 2}$ . Then $\int (12 t^{1 / 2} - 3 t) d t = 12 \cdot \frac{t^{3 / 2}}{3 / 2} - 3 \cdot \frac{t^{2}}{2} + C = 8 t^{3 / 2} - \frac{3}{2} t^{2} + C$ .

(b) $\frac{d}{d t} (8 t^{3 / 2} - \frac{3}{2} t^{2} + C) = 8 \cdot \frac{3}{2} t^{1 / 2} - 3 t + 0 = 12 \sqrt{t} - 3 t$ , which is $R (t)$ .

(c) The two models differ only by the constant, and a constant has derivative 0, so both have $W^{'} (t) = R (t)$ . Ben's model puts 150 more gallons of water in the tank than Ana's at every time $t$ . The rate alone cannot tell the two apart; the amount of water at one time, such as $t = 0$ , would.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Rewrites $12 \sqrt{t}$ as $12 t^{1 / 2}$ and antidifferentiates it to $8 t^{3 / 2}$ .
Antidifferentiates $- 3 t$ to $- \frac{3}{2} t^{2}$ and ends the indefinite integral with $+ C$ .
Differentiates the answer term by term, the constant to 0, and gets $R (t) = 12 \sqrt{t} - 3 t$ .
Says both models have derivative $R (t)$ because they differ only by a constant.
Interprets the difference as 150 more gallons in the tank at every time in Ben's model, with units.

Part 6 Check yourself A table of derivatives, used backward

The first three questions use the table. The last two stand on their own.

A student made this table of derivatives while studying. Every row is correct. Use it to answer the questions.

A student's table of derivatives
$F (x)$	$F^{'} (x)$
$\tan x$	$\sec^{2} x$
$x e^{x}$	$(x + 1) e^{x}$
$\sin (3 x)$	$3 \cos (3 x)$
$\ln (x^{2} + 1)$	$\frac{2 x}{x^{2} + 1}$

Case question 1 · Multiple choice

No calculator. Use the table. Which of these is $\int 6 \cos (3 x) d x$ ?

Case question 2 · Calculate

No calculator. Let $H$ be any antiderivative of $h (x) = 5 \sec^{2} x$ on the interval $- \frac{π}{2} < x < \frac{π}{2}$ . Use the table to find $H (\frac{π}{4}) - H (- \frac{π}{4})$ .

Your answer

Case question 3 · Decide and justify

No calculator. A classmate reads the table's second row and writes $\int x e^{x} d x = x e^{x} + C$ . Is the classmate right?

Which reason supports your answer?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student finds $\int (\frac{4}{x^{2}} - 3 \sin x) d x$ . Tap the first line with a mistake.

Check yourself · Sketch the graph

No calculator. $F$ is an antiderivative of $f (x) = x^{2} + 1$ . Sketch a possible graph of $F$ for $x$ from 0 to 4. The first point is fixed at 0.

Drag a point up or down, or tap it and use the arrow keys. Only the shape is checked.

Part 7 Keep it Bring back earlier ideas

This lesson leans on an accumulation function from earlier in the unit and on the chain rule from Unit 3. Pulling each back up is what keeps it.

Bring it back · Calculate

No calculator. Let $g (x) = \int_{1}^{x} \sqrt{t^{3} + 1} d t$ . Find $g^{'} (2)$ .

Your answer

Bring it back · Multiple choice

No calculator. If $f (x) = e^{x^{3}}$ , what is $f^{'} (x)$ ?

In three lines

An antiderivative of $f$ is any function whose derivative is $f$ , and they differ from each other by constants.
$\int f (x) d x = F (x) + C$ names them all, and the derivative rules read backward give $F$ .
Some functions have no closed-form antiderivative, and then a calculator or an accumulation function does the work.

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

In a written answer, show the antiderivative of each term, end every indefinite integral with $+ C$ , and differentiate your result to check it before you move on. When a definite integral's integrand has no closed-form antiderivative, write the integral with its limits and $d x$ , then evaluate it with the calculator.

Questions students ask

Why does every indefinite integral need

+ C

Because a function has infinitely many antiderivatives, one for each constant. $x^{2}$ , $x^{2} + 1$ and $x^{2} - 9$ all have derivative $2 x$ , and $\int 2 x d x = x^{2} + C$ names all of them.

What is the difference between

\int f (x) d x

and

\int_{a}^{b} f (x) d x

The first, with no limits, is a family of functions, $F (x) + C$ . The second, with limits, is a single number. Antiderivatives connect them: when $f$ is continuous on $[a, b]$ , the Fundamental Theorem of Calculus gives $\int_{a}^{b} f (x) d x = F (b) - F (a)$ .

Why is it

\ln | x |

and not

\ln x

$\ln x$ is only defined for $x > 0$ , but $\frac{1}{x}$ is also defined for $x < 0$ . For $x < 0$ the chain rule gives $\frac{d}{d x} \ln (- x) = \frac{1}{x}$ , so $\ln | x |$ covers both sides.

Is there a product rule or a quotient rule for antiderivatives?

Not one you can apply term by term. Rewrite first when you can, by expanding a product or splitting a fraction into powers, and check the result by differentiating. Topic 6.9 adds a method that reverses the chain rule.

If I can't find an antiderivative, does that mean there isn't one?

No. Some integrands, such as $2 x e^{x^{2}}$ , look hard and have a closed form. Others, such as $e^{x^{2}}$ , have antiderivatives with no closed form. Either way, a continuous function on an interval always has an antiderivative.

Course alignment, for teachers

AP Calculus AB topic 6.8, Unit 6: Integration and Accumulation of Change.

This lesson maps all 3 essential knowledge statements of the topic to the part that teaches each one: 3 are checked by a scored question. The map is in the lesson file.